HDU 2141 Can you find it?（二分）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 33248 Accepted Submission(s): 8241

Problem Description
Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

Input
There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

Output
For each case, firstly you have to print the case number as the form “Case d:”, then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print “YES”, otherwise print “NO”.

Sample Input
3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10


Sample Output
Case 1:
NO
YES
NO


Author
wangye


Source
HDU 2007-11 Programming Contest

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
int main()
{
    long long l,m,n,cnt=1;
    while(cin>>l>>m>>n)
    {
    long long a[510],b[510],c[510],num[250010];
    for(int i=0;i<l;i++)//输入三个数组 
    cin>>a[i];
    for(int i=0;i<m;i++)
    cin>>b[i];
    for(int i=0;i<n;i++)
    cin>>c[i];
    int k=0;
    for(int i=0;i<l;i++)
        for(int j=0;j<m;j++)
            num[k++]=a[i]+b[j];//前两个数组两两相加得到新数组 
    sort(num,num+k);//从小到大排序 
    int s;
    cin>>s;
    cout<<"Case "<<cnt++<<":"<<endl;//输出第几个测试实例 
    while(s--)
    {
        long long x;
        cin>>x;
        bool st =false;
        for(int i=0;i<n;i++)//第三个数组和新数组mid相加判断 
        {
        int l=0,r=k-1;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(num[mid]+c[i]==x)
            {
                st=true;
                break;
            }
            else if(num[mid]+c[i]<x)
                l=mid+1;
            else
                r=mid-1;
        }   
        if(st)
            break;
        }
        if(st)
        cout<<"YES"<<endl;
        else
        cout<<"NO"<<endl;
    }   
    }
    return 0;
}